Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $x \neq 0$. $r = \dfrac{-8}{40x + 72} \div \dfrac{-10}{20x^2 + 36x} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{-8}{40x + 72} \times \dfrac{20x^2 + 36x}{-10} $ When multiplying fractions, we multiply the numerators and the denominators. $r = \dfrac{ -8 \times (20x^2 + 36x) } { (40x + 72) \times -10 } $ $ r = \dfrac {-8 \times 4x(5x + 9)} {-10 \times 8(5x + 9)} $ $ r = \dfrac{-32x(5x + 9)}{-80(5x + 9)} $ We can cancel the $5x + 9$ so long as $5x + 9 \neq 0$ Therefore $x \neq -\dfrac{9}{5}$ $r = \dfrac{-32x \cancel{(5x + 9})}{-80 \cancel{(5x + 9)}} = -\dfrac{32x}{-80} = \dfrac{2x}{5} $